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Questions tagged [scissors-congruence]

Equidecomposability of polyhedra under cutting-and-pasting along faces, and generalizations. Hilbert's third problem, Dehn's invariants. Homological developments motivated by these issues.

3 votes
2 answers
450 views

Let $\text{Var}_k$ be the category of varieties over $k$. The Grothendieck ring of $\text{Var}_k$ is the ring $$K_0(\text{Var}_k)= \{\text{isomorphism classes of finite type varieties over }k \}/([X]=[...
Bockman Cheung's user avatar
2 votes
0 answers
187 views

Recall that a motivic measure is a function $\mu$ that associates to each variety $X/k$ an element of some ring $R$, such that If $X\cong X'$ then $\mu(X)=\mu(X')$. If $Y\subseteq X$ is a closed ...
BHT's user avatar
  • 211
10 votes
0 answers
240 views

Suppose $A,B$ are polygons of equal area. By the Wallace-Bolyai-Gerwien theorem, $A$ and $B$ are equidissectable: we can make finitely many straight-line cuts in $A$ and rearrange the resulting pieces ...
Noah Schweber's user avatar
6 votes
1 answer
251 views

In C-H Sah's book Hilbert's third problem: scissors congruence, the author defines the data for abstract scissors congruence in order to prove Zylev's theorem by combinatorial means in great ...
Roselyn van Lauwe's user avatar
8 votes
1 answer
930 views

$\DeclareMathOperator\SL{SL}$I am reading some papers about the third homology of linear groups. In particular for the $\SL_{2}$ over a field. Why is it important to study these homologies? I have ...
Liddo's user avatar
  • 269
3 votes
2 answers
354 views

Definition: Two finite sets of polygons $A$ and $B$ are congruent if we can match polygons in $A$ in a one-one manner with polygons in $B$ with each matched pair of polygons mutually congruent. ...
Nandakumar R's user avatar
  • 7,583
3 votes
1 answer
114 views

Given two polygons of equal area with horizontal foliations, can one describe the obstruction (if there is any but I suspect the answer to be yes) to scissor-equivalence respecting the horizontal ...
Roland Bacher's user avatar
21 votes
3 answers
1k views

Question. For which $N$ it is possible to cut a regular $N$-gon into congruent pieces such that the center of the regular polygon lies strictly inside one of the pieces? For $N=3,4$ there are trivial ...
Fedor Nilov's user avatar
2 votes
1 answer
359 views

See Wikipedia for Monsky's theorem which states: it is not possible to dissect a square into an odd number of triangles all of equal area. Questions: Are there quadrilaterals that allow partition into ...
Nandakumar R's user avatar
  • 7,583
2 votes
1 answer
334 views

It is well-known that a square can be cut into a finite number of squares all of mutually different sides (hence mutually non-congruent) - for example, see https://en.wikipedia.org/wiki/...
Nandakumar R's user avatar
  • 7,583
2 votes
1 answer
246 views

Given any integer $n$, any rectangular region or any sector of a disc (including the full disk as a boundary case) can be cut into $n$ mutually congruent pieces - by equally spaced parallel lines and ...
Nandakumar R's user avatar
  • 7,583
6 votes
1 answer
304 views

Sydler proved something of a converse to Dehn's negative resolution of Hilbert's 3rd problem. To quote Wikipedia, Sydler showed that "every two Euclidean polyhedra with the same volumes and Dehn ...
Joseph O'Rourke's user avatar
6 votes
1 answer
636 views

The following questions seem related to the still open question whether there is a point(s) whose distances from the 4 corners of a unit square are all rational. To cut a unit square into n (a finite ...
Nandakumar R's user avatar
  • 7,583
2 votes
0 answers
631 views

It is easy to see that an equilateral triangle can be cut into 2 identical 30-60-90 degrees right triangles which can then be patched together to form a 30-30-120 degrees triangle. So, via 2 ...
Nandakumar R's user avatar
  • 7,583
2 votes
1 answer
822 views

Question: Given an $N$-vertex polygon (not necessarily convex). It is to be cut into the least number of acute isosceles triangles. Based on this MathSE discussion, one can think of a method to get $\...
Nandakumar R's user avatar
  • 7,583

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